Surface area and volume of sphere - class-IX

$\displaystyle2\pi r=\frac{132}{7}\implies r=3$

$\therefore$ Capacity $\displaystyle=\frac{2}{3}\pi r^3=18\pi$

Let the sides of cube be $s$ and radius of sphere be $r$

Volume of spherical shell
$= \displaystyle \frac{4 \pi}{3} (R^3 - r^3) = \frac{4 \pi}{3} (12^3 - 10^3)$
$=\displaystyle \frac{4}{3} \times \pi \times (12 -10) (12^2 +12 \times 10 +10^2)$
$=\displaystyle \frac{4}{3} \times \pi \times 2 \times 264 cm^3$
$Weight = volume \times density = \displaystyle \frac{4}{3}\times \pi \times 364 \times 4.8 = 14.64 kg$

Given,

The radius of the sphere $=r$

The radius of the cylinder $=r$

The height of the cylinder $=2r$

We know that the volume of the sphere

${{V} _{1}}=\dfrac{4}{3}\pi {{r}^{3}}$

We know that the volume of the cylinder

$ {{V} _{2}}=\pi {{r}^{2}}h $

$ {{V} _{2}}=\pi {{r}^{2}}\left( 2r \right) $

$ {{V} _{2}}=2\pi {{r}^{3}} $

Therefore, the required ratio

$ \dfrac{{{V} _{1}}}{{{V} _{2}}}=\dfrac{\dfrac{4}{3}\pi {{r}^{3}}}{2\pi {{r}^{3}}} $

$ \dfrac{{{V} _{1}}}{{{V} _{2}}}=\dfrac{2\pi {{r}^{3}}}{3\pi {{r}^{3}}} $

$ \dfrac{{{V} _{1}}}{{{V} _{2}}}=\dfrac{2}{3} $

$ {{V} _{1}}:{{V} _{2}}=2:3 $

Hence, this is the answer.

Given, internal diameter $=8cm$ and external diameter $=10cm$
Volume of a hollow sphere of outer Radius R and inner radius r $ = \frac { 4 }{ 3 } \pi ({R}^{2} -{ r}^{ 3 }) $
Inner radius of the spherical shell $ = \frac {8}{2} = 4 cm $
Outer radius of the spherical shell $ = \frac {10}{2} = 5  cm $
Hence, volume of spherical shell $ = \frac { 4 }{ 3 } \times \pi \times ({5}^{3} - {4}^{3}) = \frac { 4 }{ 3 } \times \pi  \times 61 = \frac {244\pi}{3}  { cm }^{ 3 }  $

Volume of speed used $=$ $\dfrac { 2 }{ 3 } \pi \left( { r } _{ 1 }^{ 3 }-{ r } _{ 2 }^{ 3 } \right) $

Volume of a hollow sphere of outer Radius R and inner radius r $ =

\cfrac { 4 }{ 3 } \pi ({R}^{3} -{ r }^{ 3 }) $

Inner radius of the spherical shell $ = \cfrac {8}{2} = 4 cm $

Outer radius of the spherical shell $ = \cfrac {12}{2} = 6  cm $

Volume of a cone $ = \cfrac { 1 }{ 3 } \pi { r }^{ 2 }h $  where r is the

radius of the base of the cone and h is the height.
Radius of the cone $ = \cfrac{8}{2} = 4  cm $

Now, Volume of the hollow sphere $ = $ Volume of cone
$ => \cfrac { 4 }{ 3 } \pi ({6}^{3} -{ 4 }^{ 3 }) = \cfrac { 1 }{ 3 } \pi { 4 }^{ 2 }h $
$ => 4 \times(216-64) = 16h $
$ h = 38  cm $

$R = 10 m$$
$r  = 2 m$
Volume $=\cfrac{4}{3}\pi (R^3-r^3)$
$=\cfrac{4}{3}\pi (10^3-2^3)$

R $= 8\ cm$
r  $= 4\ cm$
Volume = $\dfrac{4}{3}\pi (R^3-r^3)$

Thickness, $T = R - r$
$10 = R - 25$
$R = 35$ cm
$r  = 25$ cm
Volume = $\cfrac{4}{3}\pi (R^3-r^3)$
$=$ $\cfrac{4}{3}\pi (35^3-25^3)$
$=$ $\cfrac{4}{3}\pi (42875-15625)$
$=$ $\cfrac{4}{3}\pi (27250)$
$=$ $\cfrac{109000 \pi}{3}$
$=$ $114,087 \space\ cm$

Thickness, $T = R - r$
$5 = 7 - r$
$r = 7 - 5 = 2$ m
$R = 7$ m
Volume = $\cfrac{4}{3}\pi (R^3-r^3)$
= $\cfrac{4}{3}\pi (7^3-2^3)$
= $\cfrac{4}{3}\pi (343-8)$
= $\cfrac{4}{3}\pi (335)$
= $\cfrac{1340 \pi}{3}$
= $1402.53\space\ m$

Outer radius, $R = 4$ cm
Inner radius, $r  = 2$ cm
Volume = $\cfrac{4}{3}\pi (R^3-r^3)$
= $\cfrac{4}{3}\pi (4^3-2^3)$
= $\cfrac{4}{3}\pi (64-8)$
= $\cfrac{4}{3}\pi (56)$
= $\cfrac{224 \pi}{3}$
= $74.666\pi $
= $234.45 \space\ cm^3$

Outer radius, $R = 14$ cm
Inner radius, $r  = 7$ cm
Volume = $\cfrac{4}{3}\pi (R^3-r^3)$
= $\cfrac{4}{3}\pi (14^3-7^3)$
= $\cfrac{4}{3}\pi (2744-343)$
= $\cfrac{4}{3}\pi (2401)$
= $\cfrac{9604 \pi}{3}$
= $3201.33\pi $
$\approx 10000 \space\ m^3$

$R = 5$ ft
$r  = 3$ ft
Volume = $\cfrac{4}{3}\pi (R^3-r^3)$
= $\cfrac{4}{3}\pi (5^3-3^3)$
= $\cfrac{4}{3}\pi (125-27)$
= $\cfrac{4}{3}\pi (98)$
= $\cfrac{392 \pi}{3}$
= $130.666\pi $
= $410.293 \space\ ft^3$

$R = 5$ cm
$r  = 2.5$ cm
Volume = $\cfrac{4}{3}\pi (R^3-r^3)$
= $\cfrac{4}{3}\pi (5^3-2.5^3)$
= $\cfrac{4}{3}\pi (125-15.625)$
= $\cfrac{4}{3}\pi (109.375)$
= $\cfrac{437.5 \pi}{3}$
= $1312.5\pi $
= $457.91 \space\ cm^3$

Outer radius, $R = 24$ cm
Inner radius, $r  = 6$ cm
Volume = $\cfrac{4}{3}\pi (R^3-r^3)$
= $\cfrac{4}{3}\pi (24^3-6^3)$
= $\cfrac{4}{3}\pi (13824-216)$
= $\cfrac{4}{3}\pi (13608)$
= $\cfrac{54432 \pi}{3}$
= $18144\pi $
= $56972 \space\ cm^3$

$R = 3.0$ in
$r  = 0.1$ in
Volume = $\cfrac{4}{3}\pi (R^3-r^3)$
= $\cfrac{4}{3}\pi (3^3-0.1^3)$
= $\cfrac{4}{3}\pi (27-0.001)$
= $\cfrac{4}{3}\pi (26.999)$
= $\cfrac{107.996 \pi}{3}$
= $35.998\pi $
= $113.035 \space\ in^3$